What is (12.5) N/(m^2)*((2m)^3)/3 ?

The solution to (12.5) N/(m^2)*((2m)^3)/3 is (100Nm)/3

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simplify (12.5) N/(m^2)*((2m)^3)/3

simplify

(12.5) \frac{N}{m ^{2}} \cdot \frac{( 2 m ) ^{3}}{3}

\frac{100 N m}{3}

Solution steps

(12.5) \frac{N}{m ^{2}} \cdot \frac{( 2 m ) ^{3}}{3}

12.5 = \frac{25}{2}

= \frac{25}{2} \cdot \frac{N}{m ^{2}} \cdot \frac{( 2 m ) ^{3}}{3}

Simplify

(2 m)^{3} : 8 m^{3}

= \frac{25}{2} \cdot \frac{N}{m ^{2}} \cdot \frac{8 m ^{3}}{3}

Apply the fraction rule:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{25 N \cdot 8 m ^{3}}{2 m ^{2} \cdot 3}

Factor the number:

8 = 2 \cdot 4

= \frac{25 N \cdot 2 \cdot 4 m ^{3}}{2 m ^{2} \cdot 3}

Cancel the common factor:

2

= \frac{25 N \cdot 4 m ^{3}}{m ^{2} \cdot 3}

Simplify

\frac{m ^{3}}{m ^{2}} : m

= \frac{25 N \cdot 4 m}{3}

Multiply the numbers:

25 \cdot 4 = 100

= \frac{100 N m}{3}

x^{3} - 2 x^{2} - 15 x

x^{3} - 9 x^{2} - x + 9 \geq 0

- 15 = - \frac{5}{8} w

\frac{x + 1}{2} = \frac{21}{14}

\frac{x + 1}{5} = \frac{x + 3}{3}

What is (12.5) N/(m^2)*((2m)^3)/3 ?
The solution to (12.5) N/(m^2)*((2m)^3)/3 is (100Nm)/3